For the first-order reaction half-life is 14 s. The time required for the initial concentration to reduce to 1/8th of its value is:

42 s

The correct answer is option 3. 42 s.

Given,

Half-life, \(t_{1/2} = 14 s\)

We know,

\(t_{1/2} = \frac{0.693}{k}\)

or, \(k = \frac{0.693}{t_{1/2}}\)

\(⇒ k =\frac{0.693}{14}\)

\(∴ k =  0.0495\)

Let,'a' be the initial concentration, then according to the question,

final concentration \(= \frac{a}{8}\)

\(∴ k = \frac{2.303}{t}log \frac{a}{a/8}\)

\(⇒ t = \frac{2.303}{k}log \frac{a}{a/8}\)

\(⇒ t = \frac{2.303}{0.0495}log 8\)

\(⇒ t = \frac{2.303}{0.0495}× 0.903\)

\(⇒ t = 42.01 s\)

\(∴ t ≈ 42 s\)